Solution 2

Let $\{p_1, p_2,p_3,\ldots,p_6\}$ be the probability of each number on dice - for both dices. Then $p_1 + p_2 +\ldots+p_6 = 1,$ and $P (doubles) = p_1^2 + p_2^2 +\ldots+p_6^2.$ By Jensen's inequality, (since $x^2$ is convex)

Thus, probability of a double is $\displaystyle \sum_i p_i^2~\geq~\frac{1}{6}.$

Solution 4

You can as well throw one die twice. let $p_x$ be the probability of getting $x$ with one die. Then $\displaystyle \sum_x p_x^2$ is minimal when the largest $p_x^2$ is minimal as $x^2$ convex, which is when $\displaystyle p_x=\frac{1}{6}$ for all $x.$ So $\displaystyle \sum_x p_x^2 \ge \frac{1}{6}$.